Lines and Angles Test

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Lines and Angles Test - 40

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Question 1
1 / -0

In figure, $$POR$$ is a line. The value of $$a$$ is

A
$$40^o$$

B
$$45^o$$

C
$$55^o$$

D
$$60^o$$

Solution

Option (a) is correct.
It has been given that $$POR$$ is a straight line.
$$\therefore \angle POQ +\angle QOR =180^o$$
$$\Rightarrow (3a+5)^o +(2a-25)^o =180^o$$
$$\Rightarrow 5a=180^o +20^o =200^o$$
$$\Rightarrow a=\dfrac{200^o}{5}$$
$$\Rightarrow a=40^o$$
Question 2
1 / -0

The difference of two complementary angles is $$30^o$$. Then, the angles are:

A
$$60^o, 30^o$$

B
$$70^o, 40^o$$

C
$$20^o, 50^o$$

D
$$105^o, 75^o$$

Solution

Let $$x$$ and $$y$$ be the angles.
Then, $$x+y=90^o ....(1)$$ [ Complementary angle]
Also, $$x-y=30^o ....(2)$$ [ Given ]
By addition of both equations, we get,
$$\Rightarrow 2x=120^o$$
$$\Rightarrow x=\dfrac{120^o}{2}$$
$$\Rightarrow x=60^o$$.
Putting the value of $$x$$ in equation $$(1)$$, we get,
$$\Rightarrow 60+y=90^o$$
$$\Rightarrow y=90^o -60^o$$
$$\Rightarrow y=30^o$$
Therefore, the required angles are $$30^o$$ and $$60^o$$.
Hence, option $$A$$ is correct.
Question 3
1 / -0

In figure, $$a$$ and $$b$$ are :

A
alternate exterior angles

B
corresponding angles

C
alternate interior angles

D
vertically opposite angles

Solution

Here, $$m$$ and $$n$$ are the straight lines and transversal $$l$$ intersects both the lines $$m$$ and $$n$$.
Therefore, the angle $$a$$ and $$b$$ are alternate angles because of their position, that is, in the opposite side of the transversal and they are both lying on inside of a region covered by line $$m$$ and $$n$$.
So, they are alternate interior angles.
Hence, option $$(C)$$ is correct.
Question 4
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In figure, $$\angle AOC$$ and $$\angle BOC$$ form a pair of

A
vertically opposite angles.

B
complementary angles.

C
alternate interior angles.

D
linear pair of angles.

Solution

In figure, $$\angle AOC$$ and $$\angle BOC$$ lie on a straight line and on the same side of the straight line.
So, they form a linear pair of supplementary angles that is $$\angle AOC+\angle BOC=180^{\circ}$$.

Hence, option $$(D)$$ is correct.
Question 5
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In which of the following figures, $$a$$ and $$b$$ are forming a pair of adjacent angles?

A

B

C

D

Solution

For adjacent angles both angles must have one vertex and one side common.
$$a$$ and $$b$$ have a common vertex and a common arm that's why they are an adjacent angles.
Hence, Option $$(D)$$ is correct.
Question 6
1 / -0

In figure, the value of $$a$$ is

A
$$20^o$$

B
$$15^o$$

C
$$5^o$$

D
$$10^o$$

Solution

Option $$(D)$$ is correct.
$$\Rightarrow \angle AOF =COD$$ [ vertically opposite angles ]
$$\therefore \angle COD =90^o$$
Now,
$$\angle BOC+\angle COD+\angle EOD=180^{\circ}\cdots\cdots[\because\text{making linear pair}]$$
$$\Rightarrow 40^o +90^o +5a=180^o$$
$$\Rightarrow 5a+130^o =180^o$$
$$\Rightarrow 5a=180^o-130^o$$
$$\Rightarrow 5a=50^o$$
$$\Rightarrow a=10^o$$
Question 7
1 / -0

In figure, lines $$PQ$$ and $$ST$$ intersect at $$O$$. If $$\angle POR=90^o$$ and $$x:y =3:2$$, then $$z$$ is equal to

A
$$126^o$$

B
$$144^o$$

C
$$136^o$$

D
$$154^o$$

Solution

We can see in the given figure, $$PQ$$ is a straight line
$$\therefore \angle POR +ROT +\angle TOQ =180^o$$
$$\Rightarrow 90^o +x+y=180^o$$
$$\Rightarrow x+y=90^o$$
$$\Rightarrow x+y=90^o$$ ...(1)
Given that, $$\dfrac xy =\dfrac 32$$
Let $$x=3K$$ and $$y=2K$$
From equation (1)
$$\Rightarrow 3K+2K=90^o$$
$$\Rightarrow K=\dfrac{90^o}{5}$$
$$\Rightarrow K=18^o$$
$$\therefore y=2\times 18^o =36^o$$
Now, $$SOT$$ is a straight line
$$\Rightarrow z+y=180^o$$
$$\Rightarrow z+36^o =180^o$$
$$\Rightarrow z=180^o-36^o$$
$$\Rightarrow z=144^o$$
Hence, Option $$(B)$$ is correct.
Question 8
1 / -0

In figure, which one of the following is not true ?

A
$$\angle 1+\angle 5=180^o$$

B
$$\angle 2+\angle 5=180^o$$

C
$$\angle 3+\angle 8=180^o$$

D
$$\angle 2+\angle 3=180^o$$

Solution

Given, $$PQ$$ parallel to $$RS$$ and line $$l$$ is a transversal.
$$\therefore \angle 2+\angle 5=180^0$$ (1).....[ Co-interior angles=supplementary angles ]
$$\angle 3+\angle 8=180^0$$ (2) ..... [ Co- interior angles=supplementary angles ]
$$\angle 1=\angle 2$$ (3)..... [ Vertically opposite angles ]
$$\therefore \angle 1+\angle 5=180^o$$ [ By equation (1) and (3)]
$$\angle 2=\angle 3$$ [ Alternate interior angles ]
However, $$\angle 2 +\angle 3=180^o$$ cannot be true.
Hence, option $$D$$ is correct.
Question 9
1 / -0

In figure, $$a=40^o$$. The value of $$b$$ is

A
$$20^o$$

B
$$24^o$$

C
$$36^o$$

D
$$120^o$$

Solution

We can clearly see that, $$5b$$ and $$2a$$ are linear angles
And $$a=40^0$$
$$\Rightarrow 5b+2a=180^o$$
$$\Rightarrow 5b+2\times 40^o =180^o$$
$$\Rightarrow 5b+80^o =180^o$$
$$\Rightarrow 5b=100^o$$
$$\Rightarrow b=\dfrac{100^o}{5}$$
$$\Rightarrow b=20^o$$
So, option $$A$$ is correct
Question 10
1 / -0

In figure, which of the following is true?

A
$$\angle 1=\angle 5$$

B
$$\angle 4=\angle 8$$

C
$$\angle 5=\angle 8$$

D
$$\angle 3=\angle 7$$

Solution

Given, $$PQ$$ parallel to $$RS$$ and line $$l$$ is a transversal.
$$\therefore \angle 5=\angle 8$$ [ Alternate interior angles ]
Hence, option $$C$$ is correct.
All rest pairs in options are supplementary angles.